Question

A particle experiences constant acceleration for $20sec$ after starting from rest. If it travels a distance $S_1$in the first $10sec$ and distance $S_2$ in the next $10sec$, then

• A. $S_2=S_1$
• B. $S_2=2S_1$
• C. $S_2=3S_1$
• D. $S_2=4S_1$

Answer 1

The correct option is C

$S_2=3S_1$

Step 1: Given

As per the given data,

Initial velocity, $\mathrm{u}=0$.

Let, $a$ be the constant acceleration of the particle.

Step 2: To determine the relationship between $S_1$and $S_2$

We know that, using Newton's second equation of motion-
$s=ut+\frac{1}{2}a{t}^2$

Total distance covered in $20s$,

$s=0+\frac{1}{2}\times a\times (20{)}^2=200a$

The distance traversed by the particle for the first $10seconds$ is given by,

$s_1=0+\frac{1}{2}\times a\times (10{)}^2=50a$

The distance traversed by the particle for the following $10seconds$ is calculated by,

$S_2=(Totaldis\tan cecoveredin20s)-(Totaldis\tan cecoveredinfirst10s)$

$S_2=200a-50a=150a$

Dividing $S_2$ by $S_1$,

$\frac{S_2}{S_1}=\frac{150a}{50a}=3$

$\Rightarrow S_2=3S_1$

Hence option C is correct, the relation between $S_1$ and $S_2$ is $S_2=3S_1$.